If the area bounded by the curve y + x 2 = 8 x and the line y = 12 is K sq. units, then the value of 3 K 10 is

Step-by-step solution

The curves are y = - x 2 + 8 x = - x x - 8 and y = 12 The intersection points are A and B where - x 2 + 8 x = 12 ⇒ x 2 - 8 x + 12 = 0 ⇒ x - 2 x - 6 = 0 ⇒ x = 2,6 i.e. A 2,12 and B ( 6,12 ) Thus, the required area is A = ∫ 2 6 - x 2 + 8 x - 12 d x = - x 3 3 2 6 + 4 x 2 2 6 - 12 x 2 6 = - 216 - 8 3 + 4 36 - 4 - 12 6 - 2 = - 208 3 + 128 - 48 = 80 - 208 3 = 32 3 sq. units